Question

Prove that: $\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7}$

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $-\frac{1}{2}$

$\cos \left(\frac{2\pi }{7}\right)+\cos \left(\frac{4\pi }{7}\right)+\cos \left(\frac{6\pi }{7}\right)$

Multiply and divide by $\sin \left(\frac{\pi }{7}\right)$

$\Rightarrow \frac{\sin \frac{\pi }{7}\cos \frac{2\pi }{7}+\sin \frac{\pi }{7}\cos \frac{4\pi }{7}+\sin \frac{\pi }{7}\cos \frac{6\pi }{7}}{\sin \frac{\pi }{7}}$

Convert from product to sum:

$=\frac{\sin (\frac{\pi }{7}-\frac{2\pi }{7})+\sin (\frac{\pi }{7}+\frac{2\pi }{7})+\sin (\frac{\pi }{7}-\frac{4\pi }{7})+\sin (\frac{\pi }{7}+\frac{4\pi }{7})+\sin (\frac{\pi }{7}-\frac{6\pi }{7})+\sin (\frac{\pi }{7}+\frac{6\pi }{7})}{2\sin \left(\frac{\pi }{7}\right)}$

$\Rightarrow \frac{\sin \left(\frac{-\pi }{7}\right)+\sin \left(\frac{3\pi }{7}\right)-\sin \left(\frac{-3\pi }{7}\right)+\sin \left(\frac{5\pi }{7}\right)+\sin \left(\frac{-5\pi }{7}\right)+\sin \left(\frac{7\pi }{7}\right)}{2\sin \left(\frac{\pi }{7}\right)}$

$\Rightarrow \frac{\sin \left(\frac{-\pi }{7}\right)+\sin \left(\pi \right)}{2\sin \left(\frac{\pi }{7}\right)}$

$=-\frac{1}{2}$.